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Prove there does not exist $x,n\in \Bbb{N}$ s.t. $x^2+35=7^n$.


So I can see that $x$ must be even, because an odd plus an odd as even and $7^n$ is going to be odd, so if there does exists and $x$ it'll be even so. $$ (2k)^2+35=7^n, \quad k\in \Bbb{Z}. $$ Then I think the next step would be finding a contradiction in $$(2k)^2=7^n-35$$ but I'm not sure how.

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    $\begingroup$ Won't $x$ have to be divisible by $7$? $\endgroup$ – Lord Shark the Unknown Nov 28 at 11:02
  • $\begingroup$ Clearly $n\ge 2\,$ so by the Lemma in the dupe $\,7^2\mid x^2+7(5)\iff 7\mid \gcd(x,5)\Rightarrow\!\Leftarrow\ \ $ $\endgroup$ – Bill Dubuque Nov 28 at 14:23
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    $\begingroup$ @The They are both special cases of the Lemma in the linked dupe - see my prior comment. We close such question as abstract dupes in order to avoid having thousands of dupe answers that differ only in specializing some parameters. $\endgroup$ – Bill Dubuque Nov 28 at 17:14
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    $\begingroup$ The accepted answer here clearly shows that one does not necessarily use the mentioned lemma. Also OP's original title shows "Prove there does not exist $x,n\in N$ s.t. $x^2+35=7^n$", which is very different from the linked one. $\endgroup$ – Jack Dec 2 at 14:19
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    $\begingroup$ @BillDubuque It seems like somebody edited the question title on the linked duplicate. Also, I generally feel weird about closing something as a duplicate of a closed question. I think a better solution here would be to create a new question about your lemma advertised as an abstract-duplicate FAQ and close this question (and the other one) as duplicates of that one. This will be easier to achieve if a moderator helps with the reopening and reclosing. Since you posted the lemma to that question, maybe you would like to make the abstract duplicate question? $\endgroup$ – Trevor Gunn Dec 5 at 20:36
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It is easy to observe from the equation that $x $ must be divisible by $7$. Seting $x = 7k$ , we get $$49k^2 +35 = 7^{n} \implies 7k^2 +5 = 7^{n-1}$$

But note that $7k^2$ and $7^{n-1} $ are divisible by $7$ but $5$ is not , which is a contradiction. Hence there is no possible value of $x,n \in \mathbb{N}$

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    $\begingroup$ Technically this is not quite complete since we must discuss the case $n=1$, although that's rather trivial since the LHS is at least $35$. $\endgroup$ – YiFan Nov 28 at 11:58
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We have that

$$x^2=7^n-35=7\cdot(7^{n-1}-5)$$

but

$$7^{n-1}-5 \equiv 2 \mod 7$$

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